Question: Is ${495963}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {495963}= &&{4}\cdot100000+ \\&&{9}\cdot10000+ \\&&{5}\cdot1000+ \\&&{9}\cdot100+ \\&&{6}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {495963}= &&{4}(99999+1)+ \\&&{9}(9999+1)+ \\&&{5}(999+1)+ \\&&{9}(99+1)+ \\&&{6}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {495963}= &&\gray{4\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {4}+{9}+{5}+{9}+{6}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${495963}$ is divisible by $9$ if ${ 4}+{9}+{5}+{9}+{6}+{3}$ is divisible by $9$ Add the digits of ${495963}$ $ {4}+{9}+{5}+{9}+{6}+{3} = {36} $ If ${36}$ is divisible by $9$ , then ${495963}$ must also be divisible by $9$ ${36}$ is divisible by $9$, therefore ${495963}$ must also be divisible by $9$.